Let’s take stock of what we know about group theory so far in the first series.

We defined a group, which is a set endowed with a binary operation satisfying 3 properties.

For each group, we considered subsets which could inherit the group structure. These were called subgroups.

Each subgroup neatly partitioned the ambient group into (left/right) cosets, which gives rise to Lagrange’s theorem (among other things).

If the subgroup is normal, then the collection of cosets forms a group, called the quotient group.

Homomorphisms are functions between groups which respect the underlying operation. The kernel (set of elements mapping to the identity) is a normal subgroup, while the image is a subgroup. By the first isomorphism theorem, every homomorphism is effectively a composition of projection G → G/N, composed with injective G/N → H.

These are basic properties of a group which seem rather natural to consider. What other interesting questions of groups can we pose?

Question 1 : is the converse to Lagrange’s theorem true?

The theorem states that a subgroup H of G satisfies |H| | |G|. Conversely, if |G| = n has a factor m, one asks if G must have a subgroup of order m.

The answer is no but finding a counter-example is no easy affair: the smallest example I can think of is of order 60, and m=30. This follows from the two observations:

any subgroup H ≤ G of index 2 is normal (because if , then G is the disjoint union of H and gH, and also the disjoint union of H and Hg; thus gH = Hg).

Thus, our naive conjecture turned out to be false. But instead of giving up, let’s consider circumstances where this is true. It turns out, if m is a prime-power which divides n = |G|, then G has a subgroup of order m. This is part of Sylow’s theorems, which offers great insight into the structure of finite groups and their subgroups.

Question 2 : how many groups, up to isomorphism, are there of order n for n = 1, 2, … ?

When n is prime, we already know the answer: the group is cyclic. With a bit of work, one can prove later that if n = p2 (p prime), then the only possibilities are Z/p2 and (Z/p) × (Z/p). Things quickly get worse as we consider higher powers of prime.

Question 3 : can we consider the structure of a group by “factoring” it into its constituents?

If N is a normal subgroup of G, then we obtain subgroup N and quotient group G/N. Conversely, one hopes to reconstruct G by looking at the smaller two constituents N and G/N. The question then becomes the following.

Question 4. Given N and H, how do we construct all groups G containing N such that ?

The extension problem is theoretically solved, but actually computing all the extensions can be a real hassle. We hope to go through some explicit examples by constructing extensions of groups of order 2n.

But even if the extension problem is solved, there remains the problem of classifying groups with no nontrivial subgroups.

Question 5. A group of order > 1 is said to be simple if its only normal subgroups are {e} and itself.

[ We exclude the trivial group since we need simple groups as building blocks for more complicated groups. In this regard, the trivial group is useless. Just like 1 is not a prime number. ]

If G is abelian, then G is simple if and only if it has prime order (Exercise: prove it, using the fact that all subgroups of abelian groups are automatically normal.) For non-abelian groups, things get much stickier. The classification of finite simple groups was a massive endeavour undertaken by numerous mathematicians, which lasted until the mid 1980s. It was only twenty years later that a gap was found and had to be filled in. Here’s a more detailed overview of the history.

Speaking of breaking a group into constituents, one naturally asks if the set of constituents is unique. Explicitly:

Question 6. Let G be a finite group. Write:

,

where each and the quotient is simple. Is the set of unique up to isomorphism?

The answer is yes: this is the Jordan-Hölder theorem, which holds in far greater generality. Its proof is rather technical so we won’t be going through it.